This is an attempt to clarify the brief derivation of the Lorentz transformation in Albert Einstein’s book [1]. Suppose that the speed of light, $ c$, is the same regardless of whether a coordinate system is or is not translating with a nonzero speed.

Next consider two coordinates systems.

The first is called $ K$ and the second is called $ K’$.

The coordinates for $ K$ are $ x,y,z,t$ and the coordinates for $ K’$ are $ x’,y’,z’,t’$.

Let $ K’$ be translating with respect to $ K$ with a speed of $ v$.

The direction of this translation of $ K’$ is along the $ x$-axis of $ K$.

Suppose light flashes at the origin of both coordinate systems at $ t=0=t’$.

The light then travels along the positive $ x$-axis, positive $ x’$-axis, negative $ x$-axis, and the negative $ x’$-axis.

In $ K$ the position of the light is $ X$, and the position of the light in $ K’$ is $ X’$.

Assume that $ t$ and $ t’$ increase from 0.

First consider the situation in $ K$.

At time $ t$, the position of the light along the positive $ x$-axis is $ X = ct.$

Similarly, the position of the light along the negative $ x$-axis is

$ X = -ct.$

These two equations give rise to

$ 0 = X – ct$

and

$ 0 = X + ct.$

A similar analysis can be done for $ K’$.

The key point is that the speed of the light must be the same regardless of whether the coordinate system is translating with a nonzero speed or not.

So it must be true that, if $ X’$ is the distance between the light and the origin of $ K’$,

$ X’ = ct’ $

and

$ X’ = -ct’ $

Equations (4) and (5) give rise to

$ 0 = X’ – ct’. $

and

$ 0 = X’ + ct’. $

Note that equations (3) and (4) can be made more general with variables $ \lambda$ and $ \beta$:

$ 0 = \lambda (X – ct) $

and

$ 0 = \beta(X + ct).$

**Since $ 0=0$, perhaps** equations (7) and (9) are equal, and **perhaps** equations (8) and (10) are equal:

$ X’ – ct’ = \lambda (X – ct) $

and

$ X’ + ct’ = \beta(X + ct) $

These equations can be added and subtracted.

Let

$ a \equiv \frac{\lambda + \beta}{2} \in \mathbb{R} $

and

$ b \equiv \frac{\lambda – \beta}{2} \in \mathbb{R}. $

Then

$ X’ = aX – bct $

and

$ ct’ = act – bX.$

**Option 1**

First consider $ t=0$.

Then equation (15) yields

$ X’ = aX.$

Next consider $ t’=0$ in $ K’$.

Then equation (16) yields

$ \frac{b c}{a} X = c^2 t = vX $

Define a velocity (see Appendix) as

$ v \equiv \frac{bc}{a} \in \mathbb{R}. $

Constrain $ v \ge 0$ to define the physical situation.

Since $ c \ge 0 $, $ b$ and $ a$ must now have the same sign.

Note that equation (17) yields

$ \Delta X’ = a \Delta X $

and if $ \Delta X’ = 1$ then

$ \frac{1}{a} = \Delta X.$

Consider for general $ t$ equation (15):

$ X’ = aX – bct $

along with the previously derived equation with $ t’=0$

$ X = \frac{c^2}{v} t \rightarrow t = \frac{v}{c^2}X $

The previous two equations yield

$ X’ = aX – b \bigg(\frac{v}{c}X \bigg) $

Suppose $ \Delta X = 1$ from the perspective of $ K’$ at $ t’=0$.

Then

$ \Delta X’ = a \bigg( 1 – \frac{v^2}{c^2}\bigg) $

By the principle of relativity the magnitudes of the displacements are equal, so using (21) and equation (25) one has:

$ \bigg| \frac{1}{a} \bigg| = \bigg| a \bigg( 1 – \frac{v^2}{c^2} \bigg) \bigg|.$

Constrain $ v\le c$ so that $ \big( 1 – \frac{v^2}{c^2}\big) \ge 0$. Then

$ a^2 = \frac{1}{1 – \frac{v^2}{c^2}}.$

Therefore,

$ a = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}$

and

$ a^2 v^2 = b^2 c^2 \rightarrow b^2 = \frac{a^2 v^2}{c^2}$

Therefore,

$ b = \pm \frac{v}{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}} $

Finally, substitute $ a$ and $ b$ into equations (15) and (16).

$ X’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}X – \pm \frac{v}{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct $

and

$ ct’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct – \pm \frac{v}{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}X. $

Choose the positive roots.

$ \boxed{ X’ = \frac{X – vt}{\sqrt{1 – \frac{v^2}{c^2}}}} $

and

$ \boxed{ t’ = \frac{t – \frac{v}{c^2} X }{\sqrt{1 – \frac{v^2}{c^2}}} }.$

**Option 2**

First consider $ t=0$. Then

$ X’ = aX$

or $ \Delta X’ = a \Delta X.$

Suppose $ \Delta X = 1$.

Then

$ \Delta X’ = a.$

Next consider $ t’=0$.

Then equation (16) yields

$ bX = act.$

Define a velocity $ v \equiv \frac{bc}{a} \in \mathbb{R}$ (see Appendix). Constrain $ v \ge 0$ to define the physical situation.

Since $ c \ge 0 $, $ b$ and $ a$ must now have the same sign.

Then

$ v X = c^2 t $

or

$ t = \frac{vX}{c^2}.$

For general $ t$ equation (15) is

$ X’ = aX – bct $

Substitute the result from $ t’=0$ to obtain

$ X’ = a X \bigg( 1 – \frac{v^2}{c^2} \bigg)$

Hence

$ \Delta X’ = a \Delta X \bigg( 1 – \frac{v^2}{c^2} \bigg)$

Suppose that $ \Delta X’ = 1$.

Then

$ 1 = a \Delta X \bigg( 1 – \frac{v^2}{c^2} \bigg)$

or

$ \frac{1}{a\big( 1 – \frac{v^2}{c^2} \big)} = \Delta X $

From the principle of relativity the magnitudes of the displacements are equal:

$ \bigg| \frac{1}{a\big( 1 – \frac{v^2}{c^2} \big)} \bigg| = |a|.$

Constrain $ v\le c$ so that $ \big( 1 – \frac{v^2}{c^2}\big) \ge 0$. Then

$ \frac{1}{ 1 – \frac{v^2}{c^2} } = a^2.$

Since

$ v^2 = \frac{b^2 c^2}{a^2} \rightarrow \frac{v^2 a^2}{c^2} = b^2 $

one has

$ b = \pm \frac{v }{c} a = \pm \frac{v }{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}.$

since, from (47), one has

$ a = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}.$

Finally, substitute $ a$ and $ b$ back into equations (15) and (16):

$ X’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}X – \pm \frac{v }{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct $

and

$ ct’ = \pm \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}ct – \pm \frac{v }{c} \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}} X.$

Choose the positive roots.

$ \boxed{ X’ = \frac{X – vt}{\sqrt{1 – \frac{v^2}{c^2}}} }$

and

$ \boxed{ t’ = \frac{t – \frac{v}{c^2} X}{\sqrt{1 – \frac{v^2}{c^2}}} }.$

**Appendix**

Recall that

$ X’ = aX – bct.$

Divide by $ a$:

$ \frac{1}{a} X’ – X = – \frac{bc}{a} t $.

Suppose $ X’=0$.

Then, since

$ X = \frac{bc}{a} t = vt $

one has

$ \frac{\Delta X}{\Delta t} = v $

Alternatively, suppose $ X=0$. Then equation (15) yields

$ X’ = -bct $

while equation (16) yields

$ \frac{1}{a}t’ = t .$

Substituting $ t = \frac{t’}{a}$ into $ X’ = -bct$ one obtains

$ X’ = -bc \frac{t’}{a} $

which leads to

$ \Delta X’ = – \frac{bc}{a} \Delta t’ $

or, with $ v \equiv \frac{bc}{a}$,

$ \frac{\Delta X’}{\Delta t’} = -v.$

**References**

[1] Albert Einstein. Relativity; *The Special and General Theory*. Appendix A.

Three Rivers Press, 1961.